Integrand size = 82, antiderivative size = 367 \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\frac {1}{3} p \text {RootSum}\left [b^3 c p^3-3 a b^2 c p^2 q+3 a^2 b c p q^2-a^3 c q^3+3 a b^2 c p^2 \text {$\#$1}^3+d p^3 \text {$\#$1}^3-6 a^2 b c p q \text {$\#$1}^3+3 a^3 c q^2 \text {$\#$1}^3+3 a^2 b c p \text {$\#$1}^6-3 a^3 c q \text {$\#$1}^6+a^3 c \text {$\#$1}^9\&,\frac {-b^2 p^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )+2 a b p q \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )-a^2 q^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )+a b p \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^3-a^2 q \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^3+2 a^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^6}{3 a b^2 c p^2 \text {$\#$1}^2+d p^3 \text {$\#$1}^2-6 a^2 b c p q \text {$\#$1}^2+3 a^3 c q^2 \text {$\#$1}^2+6 a^2 b c p \text {$\#$1}^5-6 a^3 c q \text {$\#$1}^5+3 a^3 c \text {$\#$1}^8}\&\right ] \]
Time = 0.42 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\frac {1}{3} p \text {RootSum}\left [b^3 c p^3-3 a b^2 c p^2 q+3 a^2 b c p q^2-a^3 c q^3+3 a b^2 c p^2 \text {$\#$1}^3+d p^3 \text {$\#$1}^3-6 a^2 b c p q \text {$\#$1}^3+3 a^3 c q^2 \text {$\#$1}^3+3 a^2 b c p \text {$\#$1}^6-3 a^3 c q \text {$\#$1}^6+a^3 c \text {$\#$1}^9\&,\frac {-b^2 p^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )+2 a b p q \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )-a^2 q^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right )+a b p \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^3-a^2 q \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^3+2 a^2 \log \left (\sqrt [3]{q+p x^3}-\text {$\#$1}\right ) \text {$\#$1}^6}{3 a b^2 c p^2 \text {$\#$1}^2+d p^3 \text {$\#$1}^2-6 a^2 b c p q \text {$\#$1}^2+3 a^3 c q^2 \text {$\#$1}^2+6 a^2 b c p \text {$\#$1}^5-6 a^3 c q \text {$\#$1}^5+3 a^3 c \text {$\#$1}^8}\&\right ] \]
Integrate[(x^2*(b + a*x^3)*(-(b*p) + 3*a*q + 2*a*p*x^3))/((q + p*x^3)^(2/3 )*(b^3*c + d*q + (3*a*b^2*c + d*p)*x^3 + 3*a^2*b*c*x^6 + a^3*c*x^9)),x]
(p*RootSum[b^3*c*p^3 - 3*a*b^2*c*p^2*q + 3*a^2*b*c*p*q^2 - a^3*c*q^3 + 3*a *b^2*c*p^2*#1^3 + d*p^3*#1^3 - 6*a^2*b*c*p*q*#1^3 + 3*a^3*c*q^2*#1^3 + 3*a ^2*b*c*p*#1^6 - 3*a^3*c*q*#1^6 + a^3*c*#1^9 & , (-(b^2*p^2*Log[(q + p*x^3) ^(1/3) - #1]) + 2*a*b*p*q*Log[(q + p*x^3)^(1/3) - #1] - a^2*q^2*Log[(q + p *x^3)^(1/3) - #1] + a*b*p*Log[(q + p*x^3)^(1/3) - #1]*#1^3 - a^2*q*Log[(q + p*x^3)^(1/3) - #1]*#1^3 + 2*a^2*Log[(q + p*x^3)^(1/3) - #1]*#1^6)/(3*a*b ^2*c*p^2*#1^2 + d*p^3*#1^2 - 6*a^2*b*c*p*q*#1^2 + 3*a^3*c*q^2*#1^2 + 6*a^2 *b*c*p*#1^5 - 6*a^3*c*q*#1^5 + 3*a^3*c*#1^8) & ])/3
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^2 \left (a x^3+b\right ) \left (2 a p x^3+3 a q-b p\right )}{\left (p x^3+q\right )^{2/3} \left (a^3 c x^9+3 a^2 b c x^6+x^3 \left (3 a b^2 c+d p\right )+b^3 c+d q\right )} \, dx\) |
| \(\Big \downarrow \) 7266 |
| \(\displaystyle \frac {1}{3} \int -\frac {\left (a x^3+b\right ) \left (-2 a p x^3+b p-3 a q\right )}{\left (p x^3+q\right )^{2/3} \left (a^3 c x^9+3 a^2 b c x^6+\left (3 a c b^2+d p\right ) x^3+b^3 c+d q\right )}dx^3\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{3} \int \frac {\left (a x^3+b\right ) \left (-2 a p x^3+b p-3 a q\right )}{\left (p x^3+q\right )^{2/3} \left (a^3 c x^9+3 a^2 b c x^6+\left (3 a c b^2+d p\right ) x^3+b^3 c+d q\right )}dx^3\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle -p \int \frac {\left (b p-a \left (q-x^9\right )\right ) \left (b p-a \left (2 x^9+q\right )\right )}{d p^3 x^9+b^3 c p^3-a^3 c \left (q-x^9\right )^3+3 a^2 b c p \left (q-x^9\right )^2-3 a b^2 c p^2 \left (q-x^9\right )}d\sqrt [3]{p x^3+q}\) |
| \(\Big \downarrow \) 2092 |
| \(\displaystyle -p \int \frac {\left (-2 a x^9+b p-a q\right ) \left (a x^9+b p-a q\right )}{d p^3 x^9+b^3 c p^3-a^3 c \left (q-x^9\right )^3+3 a^2 b c p \left (q-x^9\right )^2-3 a b^2 c p^2 \left (q-x^9\right )}d\sqrt [3]{p x^3+q}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -p \int \left (\frac {2 a^2 x^{18}}{-a^3 c x^{27}-3 a^2 b c p \left (1-\frac {a q}{b p}\right ) x^{18}-3 a b^2 c p^2 \left (\frac {d p}{3 a b^2 c}+1+\frac {a q (a q-2 b p)}{b^2 p^2}\right ) x^9-b^3 c p^3 \left (1-\frac {a q \left (3 b^2 p^2-3 a b q p+a^2 q^2\right )}{b^3 p^3}\right )}+\frac {a b p \left (1-\frac {a q}{b p}\right ) x^9}{-a^3 c x^{27}-3 a^2 b c p \left (1-\frac {a q}{b p}\right ) x^{18}-3 a b^2 c p^2 \left (\frac {d p}{3 a b^2 c}+1+\frac {a q (a q-2 b p)}{b^2 p^2}\right ) x^9-b^3 c p^3 \left (1-\frac {a q \left (3 b^2 p^2-3 a b q p+a^2 q^2\right )}{b^3 p^3}\right )}+\frac {2 a b p q \left (-\frac {b p}{2 a q}+1-\frac {a q}{2 b p}\right )}{-a^3 c x^{27}-3 a^2 b c p \left (1-\frac {a q}{b p}\right ) x^{18}-3 a b^2 c p^2 \left (\frac {d p}{3 a b^2 c}+1+\frac {a q (a q-2 b p)}{b^2 p^2}\right ) x^9-b^3 c p^3 \left (1-\frac {a q \left (3 b^2 p^2-3 a b q p+a^2 q^2\right )}{b^3 p^3}\right )}\right )d\sqrt [3]{p x^3+q}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -p \left (a (b p-a q) \int \frac {x^9}{-d p^3 x^9-b^3 c p^3+a^3 c \left (q-x^9\right )^3-3 a^2 b c p \left (q-x^9\right )^2-3 a b^2 c p^2 \left (x^9-q\right )}d\sqrt [3]{p x^3+q}-(b p-a q)^2 \int \frac {1}{-d p^3 x^9-b^3 c p^3+a^3 c \left (q-x^9\right )^3-3 a^2 b c p \left (q-x^9\right )^2+3 a b^2 c p^2 \left (q-x^9\right )}d\sqrt [3]{p x^3+q}+2 a^2 \int \frac {x^{18}}{-d p^3 x^9-b^3 c p^3+a^3 c \left (q-x^9\right )^3-3 a^2 b c p \left (q-x^9\right )^2-3 a b^2 c p^2 \left (x^9-q\right )}d\sqrt [3]{p x^3+q}\right )\) |
Int[(x^2*(b + a*x^3)*(-(b*p) + 3*a*q + 2*a*p*x^3))/((q + p*x^3)^(2/3)*(b^3 *c + d*q + (3*a*b^2*c + d*p)*x^3 + 3*a^2*b*c*x^6 + a^3*c*x^9)),x]
3.30.67.3.1 Defintions of rubi rules used
Int[(Px_)*(u_)^(p_.)*(z_)^(q_.), x_Symbol] :> Int[Px*ExpandToSum[z, x]^q*Ex pandToSum[u, x]^p, x] /; FreeQ[{p, q}, x] && BinomialQ[z, x] && BinomialQ[u , x] && !(BinomialMatchQ[z, x] && BinomialMatchQ[u, x])
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.22 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.57
| method | result | size |
| pseudoelliptic | \(\frac {p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{3} c \,\textit {\_Z}^{9}+\left (-3 a^{3} c q +3 a^{2} b c p \right ) \textit {\_Z}^{6}+\left (3 a^{3} c \,q^{2}-6 a^{2} b c p q +3 a \,b^{2} c \,p^{2}+d \,p^{3}\right ) \textit {\_Z}^{3}-a^{3} c \,q^{3}+3 a^{2} b c p \,q^{2}-3 a \,b^{2} c \,p^{2} q +b^{3} c \,p^{3}\right )}{\sum }\left (-\frac {\left (\left (2 \textit {\_R}^{3}+q \right ) a -b p \right ) \ln \left (\left (p \,x^{3}+q \right )^{\frac {1}{3}}-\textit {\_R} \right ) \left (\left (-\textit {\_R}^{3}+q \right ) a -b p \right )}{3 \left (c \left (-\textit {\_R}^{3}+q \right )^{2} a^{3}-2 b c p \left (-\textit {\_R}^{3}+q \right ) a^{2}+a \,b^{2} c \,p^{2}+\frac {d \,p^{3}}{3}\right ) \textit {\_R}^{2}}\right )\right )}{3}\) | \(208\) |
int(x^2*(a*x^3+b)*(2*a*p*x^3+3*a*q-b*p)/(p*x^3+q)^(2/3)/(b^3*c+d*q+(3*a*b^ 2*c+d*p)*x^3+3*a^2*b*c*x^6+a^3*c*x^9),x,method=_RETURNVERBOSE)
1/3*p*sum(-1/3*((2*_R^3+q)*a-b*p)*ln((p*x^3+q)^(1/3)-_R)*((-_R^3+q)*a-b*p) /(c*(-_R^3+q)^2*a^3-2*b*c*p*(-_R^3+q)*a^2+a*b^2*c*p^2+1/3*d*p^3)/_R^2,_R=R ootOf(a^3*c*_Z^9+(-3*a^3*c*q+3*a^2*b*c*p)*_Z^6+(3*a^3*c*q^2-6*a^2*b*c*p*q+ 3*a*b^2*c*p^2+d*p^3)*_Z^3-a^3*c*q^3+3*a^2*b*c*p*q^2-3*a*b^2*c*p^2*q+b^3*c* p^3))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.61 (sec) , antiderivative size = 606, normalized size of antiderivative = 1.65 \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} c d \sqrt {\frac {\left (-c^{2} d\right )^{\frac {1}{3}}}{d}} \log \left (\frac {2 \, a^{3} c^{2} x^{9} + 6 \, a^{2} b c^{2} x^{6} + 2 \, b^{3} c^{2} + {\left (6 \, a b^{2} c^{2} - c d p\right )} x^{3} - c d q - 3 \, {\left (a x^{3} + b\right )} {\left (p x^{3} + q\right )}^{\frac {2}{3}} \left (-c^{2} d\right )^{\frac {2}{3}} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (a^{2} x^{6} + 2 \, a b x^{3} + b^{2}\right )} {\left (p x^{3} + q\right )}^{\frac {1}{3}} \left (-c^{2} d\right )^{\frac {2}{3}} + {\left (a c d x^{3} + b c d\right )} {\left (p x^{3} + q\right )}^{\frac {2}{3}} + {\left (d p x^{3} + d q\right )} \left (-c^{2} d\right )^{\frac {1}{3}}\right )} \sqrt {\frac {\left (-c^{2} d\right )^{\frac {1}{3}}}{d}}}{a^{3} c x^{9} + 3 \, a^{2} b c x^{6} + b^{3} c + {\left (3 \, a b^{2} c + d p\right )} x^{3} + d q}\right ) + \left (-c^{2} d\right )^{\frac {2}{3}} \log \left (a^{2} c^{2} x^{6} + 2 \, a b c^{2} x^{3} + b^{2} c^{2} + {\left (a c x^{3} + b c\right )} {\left (p x^{3} + q\right )}^{\frac {1}{3}} \left (-c^{2} d\right )^{\frac {1}{3}} + {\left (p x^{3} + q\right )}^{\frac {2}{3}} \left (-c^{2} d\right )^{\frac {2}{3}}\right ) - 2 \, \left (-c^{2} d\right )^{\frac {2}{3}} \log \left (a c x^{3} + b c - {\left (p x^{3} + q\right )}^{\frac {1}{3}} \left (-c^{2} d\right )^{\frac {1}{3}}\right )}{6 \, c^{2} d}, \frac {6 \, \sqrt {\frac {1}{3}} c d \sqrt {-\frac {\left (-c^{2} d\right )^{\frac {1}{3}}}{d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (a c x^{3} + b c\right )} {\left (p x^{3} + q\right )}^{\frac {2}{3}} + {\left (p x^{3} + q\right )} \left (-c^{2} d\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-c^{2} d\right )^{\frac {1}{3}}}{d}}}{c p x^{3} + c q}\right ) + \left (-c^{2} d\right )^{\frac {2}{3}} \log \left (a^{2} c^{2} x^{6} + 2 \, a b c^{2} x^{3} + b^{2} c^{2} + {\left (a c x^{3} + b c\right )} {\left (p x^{3} + q\right )}^{\frac {1}{3}} \left (-c^{2} d\right )^{\frac {1}{3}} + {\left (p x^{3} + q\right )}^{\frac {2}{3}} \left (-c^{2} d\right )^{\frac {2}{3}}\right ) - 2 \, \left (-c^{2} d\right )^{\frac {2}{3}} \log \left (a c x^{3} + b c - {\left (p x^{3} + q\right )}^{\frac {1}{3}} \left (-c^{2} d\right )^{\frac {1}{3}}\right )}{6 \, c^{2} d}\right ] \]
integrate(x^2*(a*x^3+b)*(2*a*p*x^3+3*a*q-b*p)/(p*x^3+q)^(2/3)/(b^3*c+d*q+( 3*a*b^2*c+d*p)*x^3+3*a^2*b*c*x^6+a^3*c*x^9),x, algorithm="fricas")
[1/6*(3*sqrt(1/3)*c*d*sqrt((-c^2*d)^(1/3)/d)*log((2*a^3*c^2*x^9 + 6*a^2*b* c^2*x^6 + 2*b^3*c^2 + (6*a*b^2*c^2 - c*d*p)*x^3 - c*d*q - 3*(a*x^3 + b)*(p *x^3 + q)^(2/3)*(-c^2*d)^(2/3) + 3*sqrt(1/3)*(2*(a^2*x^6 + 2*a*b*x^3 + b^2 )*(p*x^3 + q)^(1/3)*(-c^2*d)^(2/3) + (a*c*d*x^3 + b*c*d)*(p*x^3 + q)^(2/3) + (d*p*x^3 + d*q)*(-c^2*d)^(1/3))*sqrt((-c^2*d)^(1/3)/d))/(a^3*c*x^9 + 3* a^2*b*c*x^6 + b^3*c + (3*a*b^2*c + d*p)*x^3 + d*q)) + (-c^2*d)^(2/3)*log(a ^2*c^2*x^6 + 2*a*b*c^2*x^3 + b^2*c^2 + (a*c*x^3 + b*c)*(p*x^3 + q)^(1/3)*( -c^2*d)^(1/3) + (p*x^3 + q)^(2/3)*(-c^2*d)^(2/3)) - 2*(-c^2*d)^(2/3)*log(a *c*x^3 + b*c - (p*x^3 + q)^(1/3)*(-c^2*d)^(1/3)))/(c^2*d), 1/6*(6*sqrt(1/3 )*c*d*sqrt(-(-c^2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(a*c*x^3 + b*c)*(p*x^3 + q)^(2/3) + (p*x^3 + q)*(-c^2*d)^(1/3))*sqrt(-(-c^2*d)^(1/3)/d)/(c*p*x^3 + c*q)) + (-c^2*d)^(2/3)*log(a^2*c^2*x^6 + 2*a*b*c^2*x^3 + b^2*c^2 + (a*c*x ^3 + b*c)*(p*x^3 + q)^(1/3)*(-c^2*d)^(1/3) + (p*x^3 + q)^(2/3)*(-c^2*d)^(2 /3)) - 2*(-c^2*d)^(2/3)*log(a*c*x^3 + b*c - (p*x^3 + q)^(1/3)*(-c^2*d)^(1/ 3)))/(c^2*d)]
Timed out. \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\text {Timed out} \]
integrate(x**2*(a*x**3+b)*(2*a*p*x**3+3*a*q-b*p)/(p*x**3+q)**(2/3)/(b**3*c +d*q+(3*a*b**2*c+d*p)*x**3+3*a**2*b*c*x**6+a**3*c*x**9),x)
Not integrable
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.22 \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\int { \frac {{\left (2 \, a p x^{3} - b p + 3 \, a q\right )} {\left (a x^{3} + b\right )} x^{2}}{{\left (a^{3} c x^{9} + 3 \, a^{2} b c x^{6} + b^{3} c + {\left (3 \, a b^{2} c + d p\right )} x^{3} + d q\right )} {\left (p x^{3} + q\right )}^{\frac {2}{3}}} \,d x } \]
integrate(x^2*(a*x^3+b)*(2*a*p*x^3+3*a*q-b*p)/(p*x^3+q)^(2/3)/(b^3*c+d*q+( 3*a*b^2*c+d*p)*x^3+3*a^2*b*c*x^6+a^3*c*x^9),x, algorithm="maxima")
integrate((2*a*p*x^3 - b*p + 3*a*q)*(a*x^3 + b)*x^2/((a^3*c*x^9 + 3*a^2*b* c*x^6 + b^3*c + (3*a*b^2*c + d*p)*x^3 + d*q)*(p*x^3 + q)^(2/3)), x)
Timed out. \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\text {Timed out} \]
integrate(x^2*(a*x^3+b)*(2*a*p*x^3+3*a*q-b*p)/(p*x^3+q)^(2/3)/(b^3*c+d*q+( 3*a*b^2*c+d*p)*x^3+3*a^2*b*c*x^6+a^3*c*x^9),x, algorithm="giac")
Time = 21.10 (sec) , antiderivative size = 11152, normalized size of antiderivative = 30.39 \[ \int \frac {x^2 \left (b+a x^3\right ) \left (-b p+3 a q+2 a p x^3\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+\left (3 a b^2 c+d p\right ) x^3+3 a^2 b c x^6+a^3 c x^9\right )} \, dx=\text {Too large to display} \]
int((x^2*(b + a*x^3)*(3*a*q - b*p + 2*a*p*x^3))/((q + p*x^3)^(2/3)*(d*q + b^3*c + x^3*(d*p + 3*a*b^2*c) + a^3*c*x^9 + 3*a^2*b*c*x^6)),x)
symsum(log(root(59049*a^2*b*c^6*d^6*h^9*p*q^2 - 59049*a*b^2*c^6*d^6*h^9*p^ 2*q + 19683*b^3*c^6*d^6*h^9*p^3 - 19683*a^3*c^6*d^6*h^9*q^3 + 4374*a^2*b^4 *c^5*d^4*h^6*p*q + 4374*a^2*b*c^4*d^5*h^6*p*q^2 - 4374*a*b^2*c^4*d^5*h^6*p ^2*q - 2187*a^3*b^3*c^5*d^4*h^6*q^2 - 2187*a*b^5*c^5*d^4*h^6*p^2 - 2187*a^ 3*c^4*d^5*h^6*q^3 + 1458*b^3*c^4*d^5*h^6*p^3 - 567*a^2*b^4*c^3*d^3*h^3*p*q + 81*a^2*b*c^2*d^4*h^3*p*q^2 - 81*a*b^2*c^2*d^4*h^3*p^2*q + 567*a^3*b^3*c ^3*d^3*h^3*q^2 - 81*a^3*b^6*c^4*d^2*h^3*q + 162*a*b^5*c^3*d^3*h^3*p^2 + 81 *a^2*b^7*c^4*d^2*h^3*p - 81*a^3*c^2*d^4*h^3*q^3 + 27*b^3*c^2*d^4*h^3*p^3 - 3*a^3*b^3*c*d^2*q^2 - 3*a^3*b^6*c^2*d*q - a^3*d^3*q^3 - a^3*b^9*c^3, h, k )*((q + p*x^3)^(1/3)*(19683*a^24*b^14*c^8*p^21 + 192*a^19*b^4*c^3*d^5*p^26 + 4272*a^20*b^6*c^4*d^4*p^25 + 34596*a^21*b^8*c^5*d^3*p^24 + 119745*a^22* b^10*c^6*d^2*p^23 + 413343*a^26*b^12*c^8*p^19*q^2 - 688905*a^27*b^11*c^8*p ^18*q^3 + 688905*a^28*b^10*c^8*p^17*q^4 - 413343*a^29*b^9*c^8*p^16*q^5 + 1 37781*a^30*b^8*c^8*p^15*q^6 - 19683*a^31*b^7*c^8*p^14*q^7 - 192*a^20*c^2*d ^6*p^25*q^2 - 3504*a^23*c^3*d^5*p^22*q^4 - 21060*a^26*c^4*d^4*p^19*q^6 - 4 1553*a^29*c^5*d^3*p^16*q^8 + 150174*a^23*b^12*c^7*d*p^22 - 137781*a^25*b^1 3*c^8*p^20*q - 11664*a^21*b^2*c^3*d^5*p^24*q^2 - 140292*a^22*b^4*c^4*d^4*p ^23*q^2 + 264504*a^23*b^3*c^4*d^4*p^22*q^3 - 241704*a^24*b^2*c^4*d^4*p^21* q^4 - 431751*a^23*b^6*c^5*d^3*p^22*q^2 + 1516725*a^24*b^5*c^5*d^3*p^21*q^3 - 2358207*a^25*b^4*c^5*d^3*p^20*q^4 + 2079999*a^26*b^3*c^5*d^3*p^19*q^...